What are the solutions to the equation 9z^2+31=36z ?

The solutions to the equation 9z^2+31=36z are z=(6+sqrt(5))/3 ,z=(6-sqrt(5))/3

Find the zeros of 9z^2+31=36z

The zeros of 9z^2+31=36z are z=(6+sqrt(5))/3 ,z=(6-sqrt(5))/3

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9 z^{2} + 31 = 36 z

z = \frac{6 + 5}{3}, z = \frac{6 - 5}{3}

Decimal Notation

z = 2.74535 \dots, z = 1.25464 \dots

Solution steps

9 z^{2} + 31 = 36 z

Move

36 z

to the left side

9 z^{2} + 31 - 36 z = 0

Write in the standard form

a x^{2} + b x + c = 0

9 z^{2} - 36 z + 31 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 9 \cdot 31}{2 \cdot 9}

(- 36)^{2} - 4 \cdot 9 \cdot 31 = 65

z_{1, 2} = \frac{- ( - 36 ) \pm 6 5}{2 \cdot 9}

Separate the solutions

z_{1} = \frac{- ( - 36 ) + 6 5}{2 \cdot 9}, z_{2} = \frac{- ( - 36 ) - 6 5}{2 \cdot 9}

z = \frac{- ( - 36 ) + 6 5}{2 \cdot 9} : \frac{6 + 5}{3}

z = \frac{- ( - 36 ) - 6 5}{2 \cdot 9} : \frac{6 - 5}{3}

The solutions to the quadratic equation are:

z = \frac{6 + 5}{3}, z = \frac{6 - 5}{3}

\frac{4 x - 1}{2} - \frac{2 x + 5}{3} = 0

- 3 (x + 2) \geq - 8

- x^{2} + \frac{8}{2} x = 0

5 a c^{3} - 5 a

x^{2} - 49

What are the solutions to the equation 9z^2+31=36z ?
The solutions to the equation 9z^2+31=36z are z=(6+sqrt(5))/3 ,z=(6-sqrt(5))/3
Find the zeros of 9z^2+31=36z
The zeros of 9z^2+31=36z are z=(6+sqrt(5))/3 ,z=(6-sqrt(5))/3